let f(x)=∫_0 ^∞ ((1−cos(xt^2 ))/t^2 ) e^(−xt^2 ) dt with x>0 1) find a simple form of f(x) 2) calculate ∫_0 ^∞ ((1−cos(2t^2 ))/t^2 ) e^(−3t^2 ) dt .


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 38454 by maxmathsup by imad last updated on 25/Jun/18

$l e t f (x) = \int_{0}^{\infty} \frac{1 - c o s ({x t}^{2})}{t^{2}} e^{- {x t}^{2}} d t w i t h x > 0$ $1) f i n d a s i m p l e f o r m o f f (x)$ $2) c a l c u l a t e \int_{0}^{\infty} \frac{1 - c o s (2 t^{2})}{t^{2}} e^{- 3 t^{2}} d t .$
Commented bymath khazana by abdo last updated on 26/Jun/18

$t h e Q i s f (x) = \int_{0}^{\infty} \frac{1 - c o s ({a t}^{2})}{t^{2}} e^{- {x t}^{2}} d t$
Commented byabdo.msup.com last updated on 27/Jun/18
$wehave f(x)=∫_0 ^∞ ((1−cos(at^2 ))/t^2 ) e^(−xt^2 ) dt⇒ f^′ (x)= −∫_0 ^∞ (1−cos(at^2 )e^(−xt^2 ) dt =∫_0 ^∞ cos(at^2 )e^(−xt^2 ) dt −∫_0 ^∞ e^(−xt^2 ) dt but ∫_0 ^∞ e^(−xt^2 ) dt =_((√x)t=u) ∫_0 ^∞ e^(−u^2 ) (du/(√x)) =(1/(√x))∫_0 ^∞ e^(−u^2 ) du =(π/(2(√x))) and ∫_0 ^∞ cos(at^2 )e^(−xt^2 ) dt=(1/2) ∫_(−∞) ^(+∞) cos(at^2 )e^(−xt^2 ) dt =(1/2) Re( ∫_(−∞) ^∞ e^(iat^2 −xt^2 ) dt) but ∫_(−∞) ^(+∞) e^((−x+ia)t^2 ) dt =_((√(−x+ia))t=u) ∫_(−∞) ^(+∞) e^(−u^2 ) (du/(√(−x+ia))) =((√π)/(√(−x+ia))) but −x+ia=(√(x^2 +a^2 )){((−x)/(√(x^2 +a^2 ))) +((ia)/(√(x^2 +a^2 )))} =r e^(iθ) ⇒r=(√(x^2 +a^2 )) and tanθ=−(a/x) ⇒ θ=−arctan((a/x))⇒−x+ia=r e^(−iarctan((a/x))) ⇒(√(−x+ia))=(x^2 +a^2 )^(1/4) e^(−(i/2)arctan((a/x))) ⇒ ∫_(−∞) ^(+∞) e^((−x+ia)t^2 ) dt=(√π) (x^2 +a^2 )^(−(1/4)) e^((i/2)arctan((a/x))) f^′ (x)=((√π)/2)(x^2 +a^2 )^(−(1/4)) cos((1/2)arctan((a/x)))⇒ f(x)=((√π)/2) ∫_. ^x (t^2 +a^2 )^(−(1/4)) cos((1/2)arctan((a/t)))dt +c$
$w e h a v e f (x) = \int_{0}^{\infty} \frac{1 - c o s ({a t}^{2})}{t^{2}} e^{- {x t}^{2}} d t \Rightarrow$ $f^{^{'}} (x) = - \int_{0}^{\infty} (1 - c o s ({a t}^{2}) e^{- {x t}^{2}} d t$ $= \int_{0}^{\infty} c o s ({a t}^{2}) e^{- {x t}^{2}} d t - \int_{0}^{\infty} e^{- {x t}^{2}} d t b u t$ $\int_{0}^{\infty} e^{- {x t}^{2}} d t =_{\sqrt{x} t = u} \int_{0}^{\infty} e^{- u^{2}} \frac{d u}{\sqrt{x}}$ $= \frac{1}{\sqrt{x}} \int_{0}^{\infty} e^{- u^{2}} d u = \frac{π}{2 \sqrt{x}} a n d$ $\int_{0}^{\infty} c o s ({a t}^{2}) e^{- {x t}^{2}} d t = \frac{1}{2} \int_{- \infty}^{+ \infty} c o s ({a t}^{2}) e^{- {x t}^{2}} d t$ $= \frac{1}{2} R e (\int_{- \infty}^{\infty} e^{{i a t}^{2} - {x t}^{2}} d t) b u t$ $\int_{- \infty}^{+ \infty} e^{(- x + i a) t^{2}} d t =_{\sqrt{- x + i a} t = u} \int_{- \infty}^{+ \infty} e^{- u^{2}} \frac{d u}{\sqrt{- x + i a}}$ $= \frac{\sqrt{π}}{\sqrt{- x + i a}} b u t$ $- x + i a = \sqrt{x^{2} + a^{2}} {\frac{- x}{\sqrt{x^{2} + a^{2}}} + \frac{i a}{\sqrt{x^{2} + a^{2}}}}$ $= r e^{i θ} \Rightarrow r = \sqrt{x^{2} + a^{2}} a n d t a n θ = - \frac{a}{x} \Rightarrow$ $θ = - a r c t a n (\frac{a}{x}) \Rightarrow - x + i a = r e^{- i a r c t a n (\frac{a}{x})}$ $\Rightarrow \sqrt{- x + i a} = {(x^{2} + a^{2})}^{\frac{1}{4}} e^{- \frac{i}{2} a r c t a n (\frac{a}{x})} \Rightarrow$ $\int_{- \infty}^{+ \infty} e^{(- x + i a) t^{2}} d t = \sqrt{π} {(x^{2} + a^{2})}^{- \frac{1}{4}} e^{\frac{i}{2} a r c t a n (\frac{a}{x})}$ $f^{^{'}} (x) = \frac{\sqrt{π}}{2} {(x^{2} + a^{2})}^{- \frac{1}{4}} c o s (\frac{1}{2} a r c t a n (\frac{a}{x})) \Rightarrow$ $f (x) = \frac{\sqrt{π}}{2} \int_{.}^{x} {(t^{2} + a^{2})}^{- \frac{1}{4}} c o s (\frac{1}{2} a r c t a n (\frac{a}{t})) d t$ $+ c$
Commented byabdo.msup.com last updated on 27/Jun/18
$error at the final lines f^′ (x)=((√π)/2)(x^2 +a^2 )^(−(1/4)) cos{(1/2)arctan((a/x))} −(π/(2(√x))) ⇒ f(x)=((√π)/2) ∫_. ^x (t^2 +a^2 )^(−(1/4)) cos((1/2)arctan((a/t)))dt −π(√x) +c$
$e r r o r a t t h e f i n a l l i n e s$ $f^{^{'}} (x) = \frac{\sqrt{π}}{2} {(x^{2} + a^{2})}^{- \frac{1}{4}} c o s {\frac{1}{2} a r c t a n (\frac{a}{x})}$ $- \frac{π}{2 \sqrt{x}} \Rightarrow$ $f (x) = \frac{\sqrt{π}}{2} \int_{.}^{x} {(t^{2} + a^{2})}^{- \frac{1}{4}} c o s (\frac{1}{2} a r c t a n (\frac{a}{t})) d t$ $- π \sqrt{x} + c$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum